- Posted by

In Reply to:
You're killin me... posted by **jneutron**
on August 25, 2003 at 15:56:18:

You said Cu at 400w/mk. there's gotta be a length in there somewhere..Cu is 10.2 watts/inch-degree C. (sorry about the units, they say memory is the second thing to go)I found a couple of references to this number:

4 Watts per cm per degree C, which agrees with your 10.2 watts per inch per degree. Converting to MKS should give 400W per meter per Kelvin, unless I got the conversion the wrong way round and it's 0.04 W per meter per K....

Let me think this through....

The equation is actually something like:

dQ/dt (watts) = -k dT/dx (degrees/meter)

(the whole assumes unit area, BTW...)since dT/dx is degrees/distance, I think k has to be watts per degree per distance to be dimensionally correct....

Example: a 1 degree per centimeter gradient = a 100 degree per meter gradient, so....

I GOT IT THE WRONG WAY ROUND.... the thermal conductivity if Cu is 0.04W/m/K. Damn! So what effect does that have.....

The right calc is:

1.5x10

^{-7}Watts =0.04W/mKx 10^{-6}m2 x (temp gradient)

(using 0.04W/mK for the approximate thermal conducivity of CU and Ag at 300K)So, temp gradient is 1.5/4 x 10 =

3.75K/m.Ok, now I need to go think about your Seebeck effect theory here...

Maybe tomorrow!

Peter

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Follow Ups

I think we're using the same numbers..-Commuteman17:28:12 08/25/03(1)

- AHA!!!!! So that's where the error is... -
jneutron17:46:53 08/25/03(0)